$$ \begin{aligned}\rho_{Ns}=&\frac{1}{\Xi}e^{-\alpha N-\beta E_s}\end{aligned} $$

$\Xi$ 称巨配分函数。由归一化条件得出巨配分函数的表达式

$$ \sum_N\sum_s\rho_{Ns}=\frac1\Xi\sum_N\sum_se^{-\alpha N-\beta E_s}=1\\\Xi=\sum_{N=0}^\infty\sum_se^{-\alpha N-\beta E_s} $$

可以拆成辅助函数，$Z_N=\sum_se^{-\beta E_s}$粒子数为N的

$$ \begin{aligned}{\Xi}=\sum_{N=0}^\infty e^{-\alpha N}\sum_se^{-\beta E_s}=\sum_{N=0}^\infty e^{-\alpha N}Z_N\end{aligned} $$

经典情况下：

半经典情况状态数+全同原理有$\frac{d\Omega}{N!h^{Nr}}$

$$ \begin{aligned}\rho_Nd\Omega&=\frac{1}{\Xi}e^{-\alpha N-\beta H(q,p)}\frac{d\Omega}{N!h^{Nr}}\\\Xi&=\sum_{N=0}^\infty e^{-\alpha N}\int e^{-\beta H(q,p)}\frac{d\Omega}{N!h^{Nr}}\end{aligned} $$

• 归一化一条件得出巨配分函数
• 将量子态指标$s$转化为能级指标$l$

热力学函数

粒子数可变，计算其平均值

$$ \begin{aligned} \overline{N} & = \sum_N\sum_sN\rho_{Ns}\\ \Large & = \frac1\Xi\sum_{s}Ne^{-\alpha N-\beta E_s} = \frac1\Xi\sum_N\sum_s(-\frac\partial{\partial\alpha})e^{-\alpha N-\beta E_s} = \frac1\Xi{\left(-\frac\partial{\partial\alpha}\right)}\sum_{N}\sum_{s}e^{-\alpha N-\beta E_s}\\ \overline{N} & = \frac{1}{\Xi}\bigg(-\frac{\partial}{\partial\alpha}\bigg)\Xi = -\frac{\partial\ln\Xi}{\partial\alpha} \end{aligned} $$

内能公式

$$ \begin{aligned} U & = \overline{E} \Large = \sum_N\sum_sE_s\rho_{Ns} \\ &= \Large\frac1\Xi\sum_N\sum_sE_se^{-\alpha N-\beta E_s} -\frac1\Xi\sum_N\sum_s\biggl(-\frac\partial{\partial\beta}\biggr)e^{-\alpha N-\beta E_s} = \frac1\Xi\biggl(-\frac\partial{\partial\beta}\biggr)\sum_N\sum_se^{-\alpha N-\beta E_s} \\\overline{E} & = \frac{1}{\Xi}\Bigg(-\frac{\partial}{\partial\beta}\Bigg)\Xi = -\frac{\partial\ln\Xi}{\partial\beta} \end{aligned}

$$

广义力公式

$$ \begin{aligned}\overline{Y}& \begin{aligned}=\sum_N\sum_s\frac{\partial E_s}{\partial y}\rho_{Ns}=&\frac1\Xi\sum_N\sum_s\frac{\partial E_s}{\partial y}e^{-\alpha N-\beta E_s}\end{aligned}  \\&-\frac1\Xi\sum_N\sum_s\biggl(-\frac1\beta\frac\partial{\partial y}\biggr)e^{-\alpha N-\beta E_s}=\frac1\Xi\biggl(-\frac1\beta\frac\partial{\partial y}\biggr)\sum_N\sum_se^{-\alpha N-\beta E_s}\end{aligned} $$

熵公式：

元热量,熵的全微分：$dU=TdS-pdV+\mu dn\\d\overline E=TdS+Ydy+\mu d\overline N\\dS=\frac 1 T(d\overline E-Ydy-\mu d\overline N)\to dS =\beta(d\overline E-Ydy-\frac{\alpha}{\beta} d\overline N)$

• 对应项可以验证